Erdős problem #233 — wave5n
Checked 2026-07-26. Outcome: PARTIAL, not a solution of the
asymptotic problem. The new verifiable output here is an exhaustive finite
theorem through \(N=10^7\), a current-literature update missing from the live
page summary, and two exact reductions identifying the remaining large-gap
tail.
Throughout, claims are labelled as requested:
- (a) elementary-rigorous;
- (b) rigorous modulo the explicitly named theorem (or conditional
theorem);
- (c) plausible/structural-unverified;
- (d) computational-only.
0. Mandatory live-page audit before doing mathematics
Direct datacenter access returns the expected Cloudflare wall. I fetched the
rendered page and its discussion thread through the Bright Data residential
browser, saved full-page screenshots, opened the discussion URL, and clicked
“Show 1 more comments” so that all 16 comments were actually loaded.
Live URL: <https://www.erdosproblems.com/233>
The page displayed OPEN, “0 claimed proofs for this problem,”
“Interested in collaborating: None,” and “Currently working on this problem:
None.” It also displayed “This problem looks difficult: TerenceTao” and no
tractable/formalisation workers. Thus none of the mandatory stop conditions
applied. The page says it was last edited 18 January 2026; the discussion has
later comments through 31 May 2026.
Verbatim live statement
> Let \(d_n=p_{n+1}-p_n\), where \(p_n\) is the \(n\)th prime. Prove that
> \[ > \sum_{1\leq n\leq N}d_n^2\ll N(\log N)^2. > \]
The live page lists these known results:
- Cramér [Cr36] proved \(O(N(\log N)^4)\) conditional on RH.
- Selberg [Se43], still on RH, proved
\[ \sum_{1\le n\le N}\frac{d_n^2}{n}\ll(\log N)^4. \]
- The PNT gives the matching-order lower bound
\(\sum_{n\le N}d_n^2\gg N(\log N)^2\).
- The desired result would imply
\(d_n\ll n^{1/2}\log n\) for every \(n\); the page says this is known only
under RH.
- It points to OEIS A074741 and Guy, problem A8.
These are treated as the page-supplied ground truth, as required.
All 16 live comments read
The site explicitly warns that comments are unverified. The following is a
content audit, not an endorsement:
1. Desmond Weisenberg, 19 Aug 2025: located Erdős's page 440 formulation
and Cramér's conditional argument; the page was updated.
2. Terence Tao, 12 Oct 2025: supplied Selberg's weighted RH improvement;
the page was updated.
3. Bhavya Patwa, 18 Jan 2026 07:08: posted an AI-generated proposal using
prime-free intervals and a uniform Hardy–Littlewood assumption.
4. Nat Sothanaphan, 07:26: warned that the AI conversation treated the
problem as a known exercise and was likely hiding a gap.
5. Boris Alexeev, 09:10: pointed to Heath-Brown's constant-\(2\)
conjecture and to Theorem 8 of Funkhouser–Goldston–Ledoan (FGL), whose
missing input is control of very large gaps.
6. Bhavya Patwa, 09:21: said the revised AI proof was conditional on a
uniform Hardy–Littlewood \(k\)-tuple conjecture.
7. Bhavya Patwa, 09:29: accepted that the attempted implication had not
dealt with the literature's missing tail.
8. Boris Alexeev, 09:44: observed that conditional implications can still
be useful, with appropriate care.
9. Nat Sothanaphan, 11:49: clarified that ordinary Hardy–Littlewood was
not shown to imply the problem; “HL + large gap tail” is what FGL gives.
10. old-bielefelder, 16:13: apologised for an earlier AI confrontation;
no mathematical result was claimed.
11. Terence Tao, 16:41: stressed the pointwise consequence
\(d_n\ll n^{1/2}\log n\), and that even strong prime-tuple conjectures do
not supply it without a separate RH-type input.
12. Will Sawin, 27 May 2026: described a function-field analogue which he
believes reaches only \(O(N(\log N)^3)\), up to the range where every
short interval is known to contain a prime.
13. ssssasasaaasrhs, 31 May 2026 08:51: proposed
\(O(N(\log N)^3)\) from a weak Cramér-type gap bound and
Pólya–Szegő.
14. Thomas Bloom, 10:28: asked whether the problem is reasonably viewed
as “at least as hard as RH,” while noting that no implication to RH is
known.
15. StijnC, 10:34: claimed the weak Cramér assumption makes the desired
bound immediate; the displayed sentence says \(p_n\) is bounded by a
logarithmic square, so I do not use this unverified comment.
16. Will Sawin, 12:34: agreed that no implication to RH is expected, but
that a distributional prime problem not implied by RH and with no viable
unconditional route can reasonably be called harder.
Importantly, the AI proposal is only a discussion comment: the live database
still reports 0 claimed proofs.
1. Primary-source literature check
Original and conditional history
The scan of Erdős's original paper,
“The difference of consecutive primes” (1940),
page 440, asks in prime-value notation
\[ \sum_{q_i\le x}(q_{i+1}-q_i)^2=O(x\log x). \]By the PNT this is the same scale as the live page's index formulation.
(b, using the PNT)
I verified that Cramér's paper exists as H. Cramér,
The final page of the scan gives RH-conditional square-gap consequences.
FGL Sections 4–5 reproduce the Cramér/Selberg implications in modern
notation, including Selberg's
\(\mathcal C(x)\ll x\log^3x\) and weighted
\(\mathcal S(x)\ll\log^3x\) on RH.
D. R. Heath-Brown,
page 87, explicitly conjectures
\[ \sum_{p_n\le x}d_n^2\sim 2x\log x. \](c, Heath-Brown's conjecture)
His Corollary 3 gives only \(O(x\log^2x)\) under RH plus a bounded
pair-correlation hypothesis, still one logarithm above the target.
The exact conditional obstruction
Funkhouser, Goldston and Ledoan,
arXiv:1802.07609, define
\[ \mathcal C(x)=\sum_{p_{n+1}\le x}d_n^2. \]Their Theorem 8 says, conditional on their ordinary Hardy–Littlewood
hypothesis and their Strong Hardy–Littlewood hypothesis for
\(2\le k\le4\), uniformly for tuples in
\([1,x^{1/4-\delta}]\), that
\[ \mathcal C(x) =2x\log x(1+o(1)) +O\!\left( \sum_{\substack{p_{n+1}\le x\\d_n\ge x^{1/4-\delta}}}d_n^2 \right). \tag{FGL} \]This formula was checked against the PDF, not inferred from the comment.
FGL explicitly notes that its strong tuple hypothesis cannot rule out the
remaining rare very long gaps. **(b, conditional on the stated FGL
hypotheses)**
Best unconditional full second-moment bound found
Maynard's
arXiv:1201.1787 records the earlier
Peck bound
\[ \sum_{p_n\le x}d_n^2\ll_\epsilon x^{5/4+\epsilon}. \]The arXiv record itself notes that Maynard's argument reproduced Peck's
existing result.
The material update absent from the live #233 summary is Julia Stadlmann,
“On the mean square gap between primes,” arXiv:2212.10867.
Theorem 1 of the 71-page primary PDF proves
\[ \sum_{p_n\le x}d_n^2\ll_\epsilon x^{1.23+\epsilon}. \tag{1} \]Putting \(x=p_N\), using \(p_N\ll N\log N\), and absorbing the logarithm
into an arbitrarily small power gives, for every \(\eta>0\),
\[ \boxed{\sum_{n\le N}d_n^2\ll_\eta N^{1.23+\eta}}. \tag{2} \]**(b, rigorous modulo Stadlmann's Theorem 1 and the standard explicit
upper bound for \(p_N\))**
Olli Järviniemi's simultaneous
arXiv:2212.10965 improves some
first-moment tails for gaps at least \(x^{0.45}\) or \(x^{1/2}\), but its
stated theorems do not improve (1) for the complete square sum.
I searched by the exact square-sum formula, the Maynard/Peck exponent, the
Stadlmann title and exponent, and recent large-gap papers. I found no later
primary source proving the target or improving the full \(x^{1.23+\epsilon}\)
bound. This is a reported search miss, not a claim that no unindexed or
unpublished improvement exists.
2. Exact elementary reductions
Write
\[ C_N=\sum_{n=1}^N d_n^2. \]2.1 Integer layer cake
For an integer \(h\ge1\), let
\[ A_N(h)=\#\{n\le N:d_n\ge h\}. \]Since \(g^2=\sum_{h=1}^g(2h-1)\) for every positive integer \(g\),
\[ \boxed{C_N=\sum_{h\ge1}(2h-1)A_N(h).} \tag{3} \]This is an exact finite identity. (a)
2.2 Exact prime-free-interval identity
For \(h\ge1\), define
\[ E_N(h)=\#\{m\in\mathbb Z:2\le m\le p_{N+1}-h-1,\ (m,m+h]\cap\mathbb P=\varnothing\}. \]Inside the gap \((p_n,p_{n+1})\), the valid starting integers are
\[ m=p_n,p_n+1,\ldots,p_{n+1}-h-1, \]so there are exactly \((d_n-h)_+\). Hence
\[ E_N(h)=\sum_{n\le N}(d_n-h)_+. \]Summing over \(h\) and using
\(\sum_{h=1}^{g-1}(g-h)=g(g-1)/2\) gives
\[ \boxed{ C_N=(p_{N+1}-2)+2\sum_{h\ge1}E_N(h). } \tag{4} \]Thus #233 is equivalently an \(O(N\log^2N)\) bound for the total number,
over all lengths, of these prime-free intervals; the telescoping term is
only \(O(N\log N)\). (a)
2.3 An iff dyadic tail criterion
Let \(L=\log N\), and for \(j\ge0\) let
\[ B_j(N)=\#\{n\le N:2^jL\le d_n<2^{j+1}L\}. \]The gaps below \(L\) contribute at most \(NL^2\), while each \(j\)-bucket
contributes between
\(4^jL^2B_j(N)\) and \(4^{j+1}L^2B_j(N)\). Consequently
\[ \boxed{ C_N\ll N(\log N)^2 \quad\Longleftrightarrow\quad \sum_{j\ge0}4^jB_j(N)\ll N. } \tag{5} \]The constants in the two directions differ by at most \(4\). (a)
This makes the obstruction precise: fixed-\(\lambda\) Poisson laws for
\(d_n\sim\lambda\log N\) do not give the uniform summability in (5) as
\(j\) grows. A single exceptional gap can dominate a square moment.
3. Exact computation through \(10^7\)
Standalone verifier:
Run from the repository root:
python3 runs/erdos233_wave5n_reverify.py
The clean default run took 61.4 seconds here. It uses only the Python
standard library and performs two unrelated prime enumerations:
1. a monolithic odd-only Eratosthenes sieve;
2. an independently written segmented sieve whose base primes come from
trial division.
The passes compare every checkpoint, the complete gap histogram, and an
order-sensitive digest of all \(10^7\) gaps. Both gave
gap SHA-256 33d7700aeb812181cd82a6222d27a2876a9ab0fe6afac0ce7f7a28ef83cfaab5
histogram SHA-256 64c2c2dfc368291327bd2e19756d0563e65b2fbc0c08ddc0e29e30cd8cd5526f
frozen_default_output=PASS
Exact checkpoint table
Here \(G_N=\max_{n\le N}d_n\). The ratio column is diagnostic; all integer
entries are exact.
| \(N\) | \(p_N\) | \(p_{N+1}\) | \(C_N\) | \(G_N\) | first index attaining \(G_N\) | \(C_N/[N(\log N)^2]\) |
|---:|---:|---:|---:|---:|---:|---:|
| 10 | 29 | 31 | 105 | 6 | 9 | 1.980422818622 |
| 100 | 541 | 547 | 4,089 | 18 | 99 | 1.928083072701 |
| 1,000 | 7,919 | 7,927 | 95,529 | 34 | 217 | 2.001987422647 |
| 10,000 | 104,729 | 104,743 | 1,748,249 | 72 | 3,385 | 2.060876316805 |
| 100,000 | 1,299,709 | 1,299,721 | 28,095,621 | 114 | 40,933 | 2.119665102162 |
| 1,000,000 | 15,485,863 | 15,485,867 | 408,336,929 | 154 | 325,852 | 2.139364475867 |
| 10,000,000 | 179,424,673 | 179,424,691 | 5,583,894,817 | 222 | 6,957,876 | 2.149363015038 |
The record gap in this range is
\[ p_{6,957,877}-p_{6,957,876} =122,164,969-122,164,747=222. \]Exhaustive finite inequalities
The checker does not use floating point to certify these. It proves
rational lower bounds for logarithms from
\[ \log 2=2\sum_{r\ge0}\frac{(1/3)^{2r+1}}{2r+1} \]and, after writing \(n=2^km\),
\[ \log m=2\sum_{r\ge0}\frac{z^{2r+1}}{2r+1}, \qquad z=\frac{m-1}{m+1}\in[0,1/3). \]Finite positive partial sums, rounded downward at scale \(10^{12}\), are
rigorous lower bounds. Every comparison is then an integer inequality.
The exhaustive result is
\[ \boxed{C_N\le2.23\,N(\log N)^2\quad(10\le N\le10^7),} \tag{6} \]and the sharper restricted result is
\[ \boxed{C_N\le2.15\,N(\log N)^2\quad(100\le N\le10^7).} \tag{7} \]The certifying pass reports zero failed comparisons; the second pass
independently validates the entire gap sequence and all resulting sums. (d)
For orientation only, a floating scan places the largest ratio for
\(10\le N\le10^7\) at \(N=11\), about \(2.229287107735\). For
\(100\le N\le10^7\), it places the largest at \(N=9,751,264\), about
\(2.149452589274\). These locations are not used in (6)–(7).
Independent checks of the reductions
At \(N=10^7\), the program obtains
\[ \sum_{n\le N}d_n=p_{N+1}-2=179,424,689 \]and
\[ \sum_{n\le N}\frac{d_n(d_n-1)}2=2,702,235,064. \]It therefore rechecks (4) as the exact integer equality
\[ 5,583,894,817 =179,424,689+2(2,702,235,064). \]It separately reconstructs (3) and (4) from the completed gap histogram.
Some exact tail data from that histogram are:
| \(H\) | \(\#\{n\le10^7:d_n\ge H\}\) | \(\sum_{n\le10^7,\ d_n\ge H}d_n^2\) |
|---:|---:|---:|
| 100 | 15,840 | 206,207,688 |
| 120 | 4,159 | 73,241,680 |
| 150 | 474 | 12,453,900 |
| 180 | 46 | 1,702,684 |
| 200 | 11 | 480,540 |
| 220 | 2 | 97,684 |
These are finite data only, not evidence for a uniform asymptotic theorem.
(d)
4. Exactly what remains
In prime-value notation the target is
\[ \mathcal C(x)\ll x\log x. \]The clean unconditional theorem found is only
\(\mathcal C(x)\ll_\epsilon x^{1.23+\epsilon}\). The exponent gap \(0.23\)
cannot be repaired by a better treatment of logarithms.
Under the strong tuple hypotheses in FGL, everything except
\[ T_\delta(x)= \sum_{\substack{p_{n+1}\le x\\d_n\ge x^{1/4-\delta}}}d_n^2 \]has already been evaluated. The exact missing lemma would be
\[ \boxed{T_\delta(x)\ll x\log x} \tag{8} \]for the upper-bound problem, or \(T_\delta(x)=o(x\log x)\) for the
constant-\(2\) asymptotic. Ordinary fixed-\(k\) Hardy–Littlewood information
does not provide (8), because it has no uniform control over a rare gap whose
length grows this far. (b as a deduction from FGL; (8) itself is open)
No finite extension of the computation supplies that uniformity. Runtime is
essentially linear in the number of gaps: extrapolating the measured run,
\(10^8\) gaps with the same dual audit would cost roughly \(0.2\) core-hours
and the monolithic odd sieve would need about \(1.1\) GB; \(10^9\) gaps would
be of order \(2\) core-hours and \(12\) GB. Neither finite range addresses
(8), so those heavier runs were not made. **(c, linear extrapolation from
the measured run)**
The standard machinery therefore stalls at a sharply named point, rather
than at an unspecified “probably open” barrier: it lacks a uniform
square-summable upper tail for exceptional prime-free intervals/very large
prime gaps.
PARTIAL: Exact dual-sieve verification proves \(C_N\le2.23N(\log N)^2\) for every \(10\le N\le10^7\) (and \(2.15\) for \(100\le N\le10^7\)); asymptotically the unresolved input is the large-gap tail (8), while the best full unconditional bound found is Stadlmann's \(N^{1.23+\epsilon}\).